Question 12 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

The tangent to the curve 〖y=e〗^2x at the point (0, 1) meets x-axis at:

(A)(0, 1)          (B) − 1/2, 0

(C) (2, 0)         (D) (0, 2)

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Transcript

Question 12 The tangent to the curve 〖𝑦=𝑒〗^2𝑥 at the point (0, 1) meets x-axis at: (0, 1) (B) − 1/2, 0 (C) (2, 0) (D) (0, 2) Given 〖𝑦=𝑒〗^2𝑥 Finding slope of tangent at (0, 1) 〖𝑦=𝑒〗^2𝑥 Differentiating w.r.t. 𝑥 𝒅𝒚/𝒅𝒙 = 2𝒆^𝟐𝒙 At point (0, 1) Putting 𝑥=0, 𝑦=0 in 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 〖2𝑒〗^0 𝑑𝑦/𝑑𝑥 = 2 × 1 𝒅𝒚/𝒅𝒙=𝟐 Finding equation of tangent Equation of line at (𝑥1 , 𝑦1) & having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) ∴ Equation of tangent at point (0, 1) & having slope 2 is (𝒚−𝟏) = 2 (𝒙−𝟎) (𝑦−1) = 2𝑥 𝒚=𝟐𝒙+𝟏 Now, we need to find the point where tangent meets the x-axis Since point is on the x-axis, it’s y-coordinate will be 0 ∴ Point = (𝒙,𝟎) Putting 𝑦 = 0 in (1) 0 = 2𝒙 + 1 −1 = 2𝑥 (−1)/2=𝑥 𝒙 = (−𝟏)/𝟐 Thus, required point is ((−𝟏)/𝟐, 𝟎) So, the Correct answer is (B)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.